/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 Let \(g(x)=x^{2}+3 .\) Find a fu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 51

Let \(g(x)=x^{2}+3 .\) Find a function \(f\) that produces the given composition. $$(f \circ g)(x)=x^{4}+6 x^{2}+9$$

Short Answer

Expert verified
Answer: The functions f(x) and g(x) are: $$f(x) = x^2 + 5x + 6$$ $$g(x) = x^2 + 3$$

Step by step solution

01

Identify the given composition and function g(x)

The given composition is \((f \circ g)(x) = x^4 + 6x^2 + 9\), and the function g(x) is given as \(g(x) = x^2 + 3\).
02

Substitute g(x) into the composition and find f(y)

Now we need to substitute g(x) in place of y, to get: $$(f \circ g)(x) = f(g(x)) = f(x^2 + 3) = x^4 + 6x^2 + 9$$ Now we have to find the function f(y) that will produce the given composition with the input \(y = x^2 + 3\).
03

Find the function f(y) based on the given composition and g(x) substitution

Our goal is to find the function f(y) that gives us \(f(y) = x^4 + 6x^2 + 9\) when \(y = x^2 +3\). To do that, we can look at the differences between f(y) and the given composition: $$f(y) = x^4 + 6x^2 + 9 - (x^2 + 3) \Rightarrow f(y) = x^4 + 6x^2 - x^2 + 9 - 3$$ Now we can simplify this expression: $$f(y) = x^4 + 5x^2 + 6$$
04

Write the final expressions for f(x) and g(x)

So, we have found the function f(y) that produces the given composition with g(x): $$f(y) = y^2 + 5y + 6$$ And we were given the function g(x): $$g(x) = x^2 + 3$$ Therefore, the functions f(x) and g(x) that produce the given composition \((f \circ g)(x) = x^4 + 6x^2 + 9\) are: $$f(x) = x^2 + 5x + 6$$ $$g(x) = x^2 + 3$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

a. Find the linear function \(C=f(F)\) that gives the reading on the Celsius temperature scale corresponding to a reading on the Fahrenheit scale. Use the facts that \(C=0\) when \(F=32\) (freezing point) and \(C=100\) when \(F=212\) (boiling point). b. At what temperature are the Celsius and Fahrenheit readings equal?

Graphing sine and cosine functions Beginning with the graphs of \(y=\sin x\) or \(y=\cos x,\) use shifting and scaling transformations to sketch the graph of the following functions. Use a graphing utility to check your work. $$p(x)=3 \sin (2 x-\pi / 3)+1$$

Pole in a corner A pole of length \(L\) is carried horizontally around a corner where a 3 -ft-wide hallway meets a 4 -ft-wide hallway. For \(0<\theta<\pi / 2,\) find the relationship between \(L\) and \(\theta\) at the moment when the pole simultaneously touches both walls and the corner \(P .\) Estimate \(\theta\) when \(L=10 \mathrm{ft}.\)

Right-triangle relationships Draw a right triangle to simplify the given expressions. Assume \(x>0.\) $$\sin ^{-1}(\cos \theta), \text { for } 0 \leq \theta \leq \pi / 2$$

Assume that \(b>0\) and \(b \neq 1 .\) Show that \(\log _{1 / b} x=-\log _{b} x\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.